Question

Use the standard reduction potentials located in the Tables' linked above to calculate the equilibrium constant for the reaction:

Cd^2+(aq) + H2(g) ----------->Cd(s) + 2H+(aq)

Answer 1

3.1 * 10^-14

Step-by-step explanation:

Note that E°cell = 0.0592/n log K

We can obtain E°cell from the standard reduction potentials of cadmium and hydrogen

Anode reaction

H2(g) ----> 2H+ + 2e

Cathode reaction

Cd^2+(aq) + 2e -----> Cd(s)

E°cell = E°cathode - E°anode

E°cathode = –0.40 V

E°anode = 0 V

E°cell = –0.40 V - 0 V

E°cell = –0.40 V

E°cell = 0.0592/n log K

Where n=2 electrons transferred

–0.40 = 0.0592/2 log K

–0.40 = 0.0296 log K

log K = –0.40/0.0296

log K = -13.5135

K = Antilog ( -13.5135)

K = 3.1 * 10^-14